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[corrected $-$ see edit history for previous attempt]

The conjecture is false: there are infinitely many "Pell" parametrizations, some with larger values of $s/r$. For example, $$ (r,s) = (307470495089672071303, \, 295528756570432706202) $$ has $s/r \sim .961$.

This was obtained as follows. Recall that $4r^4 + 1$ factors as $(2r^2-2r+1) (2r^2+2r+1)$. Start from the first solution $(r,s) = (3,2)$, with $2r^2-2r+1 = 13$ and $2r^2+2r+1 = 5^2$. Instead of generalizing to $2r^2+2r+1 = y^2$, we generalize to $2r^2-2r+1 = 13y^2$ and $2r^2+2r+1 \equiv 0 \bmod 25$. This is a Fermat-Pell equation with a congruence condition, and since we have one solution $(r,y) = (3,2)$ there must be infinitely many others. The equation is $x^2-26y^1=1$ with $x=2r-1$, which must be positive and $1 \bmod 4$ to satisfy the sign and parity conditions on $n$. The general solution is $x + \sqrt{26} \, y = (5 + \sqrt{26})^{4k+1}$ ($k=0,1,2,\ldots$), and then the ${}\bmod 25$ condition gives $5|k$. The solution displayed above comes from $k=5$.

We can obtain further infinite families by iterating trick of switching between the $2r^2-2r+1$ and $2r^2+2r+1$ factors, and by starting from some other solution such as the $r=2679$ "outlier" or any other solution that a numerical search might find.

[corrected $-$ see edit history for previous attempt]

The conjecture is false: there are infinitely many "Pell" parametrizations, some with larger values of $s/r$. For example, $$ (r,s) = (307470495089672071303, \, 295528756570432706202) $$ has $s/r \sim .961$.

This was obtained as follows. Recall that $4r^4 + 1$ factors as $(2r^2-2r+1) (2r^2+2r+1)$. Start from the first solution $(r,s) = (3,2)$, with $2r^2-2r+1 = 13$ and $2r^2+2r+1 = 5^2$. Instead of generalizing to $2r^2+2r+1 = y^2$, we generalize to $2r^2-2r+1 = 13y^2$ and $2r^2+2r+1 \equiv 0 \bmod 25$. This is a Fermat-Pell equation with a congruence condition, and since we have one solution $(r,y) = (3,2)$ there must be infinitely many others. The equation is $x^2-26y^2=1$ with $x=2r-1$, which must be positive and $1 \bmod 4$ to satisfy the sign and parity conditions on $n$. The general solution is $x + \sqrt{26} \, y = (5 + \sqrt{26})^{4k+1}$ ($k=0,1,2,\ldots$), and then the ${}\bmod 25$ condition gives $5|k$. The solution displayed above comes from $k=5$.

We can obtain further infinite families by iterating trick of switching between the $2r^2-2r+1$ and $2r^2+2r+1$ factors, and by starting from some other solution such as the $r=2679$ "outlier" or any other solution that a numerical search might find.

[NOTE: The answer below is not correct as it stands because it does not account for the condition $r>s$. Still I do not delete it (not even for a Disciplined badge...), because the basic idea should still work: eventually there must be factors $f^2,g^2$ of $2r^2 \pm 2r + 1$ that let us multiply $r+s$ by $f/g$ to obtain an integer less than $2r$ but still with $s/r$ arbitrarily close to $1$.]

The conjecture fails infinitely often, and the ratio $s/r$ can get arbitrarily large. The easiest way to find examples is to use the same Pell family, for which one of the factors $2r^2 \pm 2r + 1$ of $4r^4+1$ is a square, and find a case where the other factor is not squarefree. Factors of $5^2$ occur often, for example when $r = 6238626641379$, when in addition to the expected $s=2584123765442$ one can take $s = 37875125392726 > 6r$; going somewhat further, $$ r = 577603898440357173330156303836731679 $$ has an extra square factor of $(5 \cdot 113)^2$, so there are three larger values of $s$ such as $$ 460945621874027221114810336774658637686 > 798r. $$ There must also be further examples such as $r=2679$ where neither of $2r^2 \pm 2r + 1$ is a square but one of them is (for example) $5$ times a square and the other has a nontrivial square factor.

[Added later] For example $(r,s) = (121644898, 878509959)$ would work with $s>7r$ (thanks to a factor of $13^2$ in $2r^2-2r+1$) except that it does not satisfy the parity condition $(r,s) \equiv (1,0) \bmod 2$.

[corrected $-$ see edit history for previous attempt]

The conjecture is false: there are infinitely many "Pell" parametrizations, some with larger values of $s/r$. For example, $$ (r,s) = (307470495089672071303, \, 295528756570432706202) $$ has $s/r \sim .961$.

This was obtained as follows. Recall that $4r^4 + 1$ factors as $(2r^2-2r+1) (2r^2+2r+1)$. Start from the first solution $(r,s) = (3,2)$, with $2r^2-2r+1 = 13$ and $2r^2+2r+1 = 5^2$. Instead of generalizing to $2r^2+2r+1 = y^2$, we generalize to $2r^2-2r+1 = 13y^2$ and $2r^2+2r+1 \equiv 0 \bmod 25$. This is a Fermat-Pell equation with a congruence condition, and since we have one solution $(r,y) = (3,2)$ there must be infinitely many others. The equation is $x^2-26y^1=1$ with $x=2r-1$, which must be positive and $1 \bmod 4$ to satisfy the sign and parity conditions on $n$. The general solution is $x + \sqrt{26} \, y = (5 + \sqrt{26})^{4k+1}$ ($k=0,1,2,\ldots$), and then the ${}\bmod 25$ condition gives $5|k$. The solution displayed above comes from $k=5$.

We can obtain further infinite families by iterating trick of switching between the $2r^2-2r+1$ and $2r^2+2r+1$ factors, and by starting from some other solution such as the $r=2679$ "outlier" or any other solution that a numerical search might find.

The conjecture fails infinitely often, and the ratio $s/r$ can get arbitrarily large. The easiest way to find examples is to use the same Pell family, for which one of the factors $2r^2 \pm 2r + 1$ of $4r^4+1$ is a square, and find a case where the other factor is not squarefree. Factors of $5^2$ occur often, for example when $r = 6238626641379$, when in addition to the expected $s=2584123765442$ one can take $s = 37875125392726 > 6r$; going somewhat further, $$ r = 577603898440357173330156303836731679 $$ has an extra square factor of $(5 \cdot 113)^2$, so there are three larger values of $s$ such as $$ 460945621874027221114810336774658637686 > 798r. $$ There must also be further examples such as $r=2679$ where neither of $2r^2 \pm 2r + 1$ is a square but one of them is (for example) $5$ times a square and the other has a nontrivial square factor.

[Added later] For example $(r,s) = (121644898, 878509959)$ with $s>7r$ (thanks to a factor of $13^2$ in $2r^2-2r+1$).

[NOTE: The answer below is not correct as it stands because it does not account for the condition $r>s$. Still I do not delete it (not even for a Disciplined badge...), because the basic idea should still work: eventually there must be factors $f^2,g^2$ of $2r^2 \pm 2r + 1$ that let us multiply $r+s$ by $f/g$ to obtain an integer less than $2r$ but still with $s/r$ arbitrarily close to $1$.]

The conjecture fails infinitely often, and the ratio $s/r$ can get arbitrarily large. The easiest way to find examples is to use the same Pell family, for which one of the factors $2r^2 \pm 2r + 1$ of $4r^4+1$ is a square, and find a case where the other factor is not squarefree. Factors of $5^2$ occur often, for example when $r = 6238626641379$, when in addition to the expected $s=2584123765442$ one can take $s = 37875125392726 > 6r$; going somewhat further, $$ r = 577603898440357173330156303836731679 $$ has an extra square factor of $(5 \cdot 113)^2$, so there are three larger values of $s$ such as $$ 460945621874027221114810336774658637686 > 798r. $$ There must also be further examples such as $r=2679$ where neither of $2r^2 \pm 2r + 1$ is a square but one of them is (for example) $5$ times a square and the other has a nontrivial square factor.

[Added later] For example $(r,s) = (121644898, 878509959)$ would work with $s>7r$ (thanks to a factor of $13^2$ in $2r^2-2r+1$) except that it does not satisfy the parity condition $(r,s) \equiv (1,0) \bmod 2$.